## Calculus: Early Transcendentals 8th Edition

Consider $\int^{\infty}_{-\infty}2xdx$ Solve the integral. $\int^{\infty}_{-\infty}2xdx=[x^{2}]_{-\infty}^{\infty}$ $=\infty-\infty$ =Indeterminate Now, $\lim\limits_{t \to \infty}\int^{t}_{-t}2xdx=\lim\limits_{t \to \infty}[x^{2}]_{-\infty}^{\infty}$ $=\lim\limits_{t \to \infty}0$ $=0$ Therefore, $\int^{\infty}_{-\infty}2xdx\ne \lim\limits_{t \to \infty}\int^{t}_{-t}2xdx$ Hence, the given statement is false.